Given fx 3x2 2x 4 find a vertex b min or max c vertex form

Given f(x)= -3x² - 2x + 4 find:
a) vertex
b) min or max
c) vertex form
d) domain
e) range
f) axis of symmetry
g) roots
h) sketch

I have done this problem over and over and keep getting crazy numbers. All help is much appreciated. Thanks!

Solution

f(x)=-3x^2-2x+4

For vertex we use vertex formula which is ,x=-b/2a

here b=-2 and a=-3

therefore x=-(-2)/2(-3)=-1/3

f(-1/3)=-3(-1/3)^2 -2(-1/3)+4=13/3

Therefore

a.vertex =(-1/3,13/3)

b. here a=-ve

so its a minimum function

c.f(x)=-3x^2-2x+4

=-3(x^2+2/3x)+4

=-3(x^2+2/3x+4/36-4/36) +4(since 2/3/2=2/6 and (2/6)^2=4/36)

=-3((x+2/6)^2-4/36)+4

=-3((x+1/3)^2-4/36)+4

=-3(x+1/3)^2 +4/12+4

y=-3(x-1/3)^2+13/3

d.domain

Here therei sno restriction on x values

Therefore domain is all real numbers

that is (-infinity,infinity)

e. range

The value of y cant be more than 13/3

Therefore range is (-infinity,13/3]

f.axis of symmetry

it is equal to x = x coordinate of vertex

x=-1/3